Question

A 20 km long, 230 kV aluminum transmission line delivers 33 MW to a city.

A) If we assume a solid cylindrical cable, what minimum diameter is needed if the voltage decrease along this run is to be no more than 1.0 % of the transmission voltage? The resistivity of aluminum is 2.7×10−8Ω⋅m.

Answer 1

here we want to find the diameter of the wire,

given data :

length of wire = 20 km

voltage , V = 230 kV

V = 230 X 10³ volt  

power , P = 33 MW

P= 33 x 10⁶ W

A resistivity of wire,

ρ = 2.7 x 10^-8 Ω⋅m

solution :

we have formula for power,

P = VI

I = P/V

I = 33 x 10⁶ / 230 X 10³

I = 0.143478 X 10³

I = 143.478 A

here we have to find resistance according to change in voltage,

R = Δ V / I

=(change in % x V) / I

=(0.01 x 230 x 10³) / 143.478

R= 16.03 Ω

Then we use equation of resistivity for finding diameter of wire ,

^ρ = R A / L

A =  ρ x L / R

= 2.7 x 10^-8 x 20000 / 16.03 { 20 km = 20000 m }

A = 33.6868 x 10^-6

π r² = 33.6868 x 10^-6

r² = 33.6868 x 10^-6 / π

   r² = 10.7186 x 10^-6

r = ( 10.7186 x 10^-6)^1/2

r = 3.2739 x 10^-3 m

diameter of wire = 2 r

= 2 x 3.2739 x 10^-3

D = 6.5458 x 10^-3 m

diameter of wire,  D = 6.5458 x 10^-3 m

diameter of wire , D = 6.54 mm { 10^-3​​​​​​​ m = 1 mm }